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Why can a torsion-free abelian group $A$ be considered as a $\mathbb{Q}$ vector space?

The author in the text I am reading says we can view $A$ as a $\mathbb{Q}$ vector space due to the embedding $A \hookrightarrow A \otimes_{\mathbb{Z}} \mathbb{Q}$. But I don't see why it follows from this.

darij grinberg
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user93826
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    Prove the tensor $;A\otimes\Bbb Q;$ is a rational vector space and, in the way to do so, pay attention to the fac that requiring $;A;$ is torsion free is of crucial importance... – DonAntonio Oct 03 '13 at 16:14
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    If the author really said that, then it was a pretty boneheaded error. Author is right that there's an embedding. – user43208 Oct 03 '13 at 16:26
  • http://www.math.harvard.edu/~mtchan/clusters.pdf [Top of page 3] Can you glean what they meant by it? – user93826 Oct 03 '13 at 16:28
  • I think the author has a rather basic mistake there and in fact, without having dived deep into the paper, I don't think he needs $;A;$ to be a $;\Bbb Q-$space (which it isn't as remarked by user####). I think he only needs to look at the span of $;A;$ within the tensor product and then pick a basis and etc. – DonAntonio Oct 03 '13 at 16:37
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    The author never states that $A$ is a $\mathbb Q$-vector space. She states that [without loss of generality] we may assume that $A$ is a rational vector space. In other words, she is saying that proving the theorem for rational vectors spaces implies the theorem for general torsion free abelian groups. – Cheerful Parsnip Oct 03 '13 at 16:41
  • @GrumpyParsnip is right: the author reduces a statement about torsion-free abelian groups to one about $Q$-vector spaces. –  Oct 03 '13 at 16:43
  • Perhaps @GrumpyParsnip, yet not remarking that the embedding is as abelian groups and remarking the rational space thing clearly is misleading and, perhaps again, even wrong. – DonAntonio Oct 03 '13 at 16:44
  • The author also does not make any WLOG assumption... – DonAntonio Oct 03 '13 at 16:44
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    I don't know -- I'd go with Grumpy Parsnips' more charitable interpretation. What the author means, I think it's pretty clear, is that it suffices to linearly order the rational vector space, since then any subgroup will inherit the order by restriction. – user43208 Oct 03 '13 at 16:47
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    @DonAntonio: The phrase "We may assume" means that we are making an additional assumption. It doesn't mean, "We have shown." – Cheerful Parsnip Oct 03 '13 at 16:48
  • Again, @GrumpyParsnip: why can we so assume? That's the important question here, imo. After all, the author later chooses a basis and etc. , so the assumption has a rather heavy importance, apparently. – DonAntonio Oct 03 '13 at 16:50
  • @DonAntonio For an explanation, see the last comment by Keenan Kidwell below his answer (or see my last comment above). I think it's pretty clear that's what the author meant. – user43208 Oct 03 '13 at 16:57
  • No, I don't agree it's pretty clear at all and, in fact, I find it pretty confusing. The problem would be easily solved, imo, just remarking that the embedding is as abelian group and then talking about Span$;A;$ ... – DonAntonio Oct 03 '13 at 16:59
  • Well, as long as everybody understands now, that's the important thing. Of course, you could write the author to complain... – user43208 Oct 03 '13 at 17:01
  • Thanks for your comments! It is not clear to me why the linear ordering specified by the author is a total ordering. If $a = v_1$ and $b = -v_1$ then neither $ba^{-1}$ nor $ab^{-1}$ are positive. – user93826 Oct 03 '13 at 20:06
  • Well, $ab^{-1}$ (or $a-b$ since we have switched to additive notation) would be positive since $a - b = 2v_1$ and $2 > 0$. But I find it easier to think of this ordering as essentially lexicographic ordering: given a linear ordering of a basis, and given two elements written as finite linear combinations of basis elements, we look for the first basis element where the rational coefficients differ, and which of the two rational coefficients is lesser (or "comes first") -- on that basis we decide which of the two elements is lesser. – user43208 Oct 05 '13 at 01:44
  • @DonAntonio: " in the way to do so, pay attention to the fac that requiring A is torsion free is of crucial importance..." It is not. If $M$ is some $R$-module and $S$ is some $R$-algebra, then $M \otimes_R S$ is an $S$-module. No assumption on $M$ is necessary. – Martin Brandenburg Oct 28 '13 at 11:32
  • Martin, I don't even remember the question, leave alone what your comment is referring to...sorry. – DonAntonio Oct 28 '13 at 11:40
  • In the following paper, https://arxiv.org/pdf/math/0501166.pdf, it is claimed that the torsion abelian groups form a Serre subcategory in the category of abelian groups, and moreover that this quotient is naturally equivalent to the category of $Q$-modules. This is on pp. 18-19. I can't see where the error is, but surely there must be one, judging from these answers. Can anyone help me elucidate the situation for myself? – Samantha Y Mar 13 '17 at 21:54

3 Answers3

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A torsion-free abelian group cannot generally be considered as a $\mathbf{Q}$-vector space. A torsion-free abelian group naturally injects into a $\mathbf{Q}$-vector space.

If $M$ is any abelian group, $M\otimes_\mathbf{Z}\mathbf{Q}$ is a $\mathbf{Q}$-vector space with a canonical abelian group homomorphism $M\rightarrow M\otimes_\mathbf{Z}\mathbf{Q}$ whose kernel is precisely the torsion subgroup of $M$.

But if, e.g., $M$ is finitely generated over $\mathbf{Z}$, it will not be a $\mathbf{Q}$-vector space and there is no way to ``regard it" as one. It embeds into a canonical one canonically as a subgroup but not a subspace.

Keenan Kidwell
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  • Thanks for your answer. What does the author mean by 'Note that A embeds into a Q-vector space via A ֒→ A ⊗Z Q, so we may assume A is a Q-vector space.' then? Have they just made an error? To give some context, they are trying to show that any torsion free abelian group can be ordered. (Group ordering that is.) – user93826 Oct 03 '13 at 16:24
  • http://www.math.harvard.edu/~mtchan/clusters.pdf [Top of page 3] – user93826 Oct 03 '13 at 16:27
  • Dear @user93826, Well, if you can order any $\mathbf{Q}$-vector space, you can order any subset of the vector space, right? So since a torsion-free abelian group is canonically a subgroup of a $\mathbf{Q}$-vector space, it would be enough to prove that a $\mathbf{Q}$-vector space can be ordered. – Keenan Kidwell Oct 03 '13 at 16:51
  • Why is $M\otimes_{\mathbf Z}\mathbf Q$ a $\mathbf Q$-vector space? I guess the action is defined on pure tensors by $r\cdot(m\otimes q):=m\otimes(rq)$ and then extended linearly, but why is this extension well-defined? – R Los Jun 30 '22 at 03:19
  • @RLos, This is the standard construction of base change for modules. For each $r\in\mathbf{Q}$, the function $\varphi_r:M\times\mathbf{Q}\to M\otimes_\mathbf{Z}\mathbf{Q}$ given by $\varphi_r(m,s)=m\otimes rs$ is $\mathbf{Z}$-bilinear. So there is a unique $\mathbf{Z}$-linear map $M\otimes_\mathbf{Z}\mathbf{Q}\to M\otimes_\mathbf{Z}\mathbf{Q}$ sending a simple tensor $m\otimes s$ to $m\otimes rs$. – Keenan Kidwell Jul 01 '22 at 15:34
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$\Bbb{Z}$ is a torsion-free abelian group but it is certainly not a $\Bbb{Q}$-vector space!

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An abelian group can be considered a $\mathbb{Q}$-vector space (in exactly one way) if and only if it is torsionfree and divisible.

Did the author of that text really say that??

user43208
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  • math.harvard.edu/~mtchan/clusters.pdf [Top of page 3] I'd appreciate if you could guide me on how the argument can be recovered. Thanks! – user93826 Oct 03 '13 at 16:30